Solve the system. 2x+y = 3 -2y = 14 - 6x
2 answers:
For this case we have the following system of equations:
We clear "y" from the second equation:
We substitute in the first equation:
So:
The solution is:
Answer:
Answer:
The solution is: (2, -1)
Step-by-step explanation:
First we rewrite the second system equation
→
Now we have the following system of equations:
To solve the system multiply the first equation by -3 and add it to the second equation
--------------------------------------
Now we substitute the value of y in the first equation and solve for x
The solution is: (2, -1)
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The number of months until the insect population reaches 40 thousand is 14.29 months and the limiting factor on the insect population as time progresses is 250 thousands.
Given that population P(t) (in thousands) of insects in t months after being transplanted is P(t)=(50(1+0.05t))/(2+0.01t).
(a) Firstly, we will find the number of months until the insect population reaches 40 thousand by equating the given population expression with 40, we get
P(t)=40
(50(1+0.05t))/(2+0.01t)=40
Cross multiply both sides, we get
50(1+0.05t)=40(2+0.01t)
Apply the distributive property a(b+c)=ab+ac, we get
50+2.5t=80+0.4t
Subtract 0.4t and 50 from both sides, we get
50+2.5t-0.4t-50=80+0.4t-0.4t-50
2.1t=30
Divide both sides with 2.1, we get
t=14.29 months
(b) Now, we will find the limiting factor on the insect population as time progresses by taking limit on both sides with t→∞, we get
(c) Further, we will sketch the graph of the function using the window 0≤t≤700 and 0≤p(t)≤700 as shown in the figure.
Hence, when the population P(t) (in thousands) of insects in t months after being transplanted by P(t)=(50(1+0.05t))/(2+0.01t) then the number of months until the insect population reaches 40 thousand 14.29 months and the limiting factor on the insect population is 250 thousand and the graph is shown in the figure.
Learn more about limiting factor from here brainly.com/question/18415071.
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